HDU1056--HangOver

HangOver
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7724 Accepted Submission(s): 3140


Problem Description
How far can you make a stack of cards overhang a table? If you have one card, you can create a maximum overhang of half a card length. (We're assuming that the cards must be perpendicular to the table.) With two cards you can make the top card overhang the bottom one by half a card length, and the bottom one overhang the table by a third of a card length, for a total maximum overhang of 1/2 + 1/3 = 5/6 card lengths. In general you can make n cards overhang by 1/2 + 1/3 + 1/4 + ... + 1/(n + 1) card lengths, where the top card overhangs the second by 1/2, the second overhangs tha third by 1/3, the third overhangs the fourth by 1/4, etc., and the bottom card overhangs the table by 1/(n + 1). This is illustrated in the figure below.



The input consists of one or more test cases, followed by a line containing the number 0.00 that signals the end of the input. Each test case is a single line containing a positive floating-point number c whose value is at least 0.01 and at most 5.20; c will contain exactly three digits.

For each test case, output the minimum number of cards necessary to achieve an overhang of at least c card lengths. Use the exact output format shown in the examples.


Sample Input
1.00
3.71
0.04
5.19
0.00


Sample Output
3 card(s)
61 card(s)
1 card(s)
273 card(s)


Source
Mid-Central USA 2001

#include<cstdio>
#include<cmath>
 
int main()
{
    double x;
    while(scanf("%lf",&x)!=EOF)
    {
        if(x==0.00)break;
        double sum=0;
        int n;
        for(n=1;;n++)
        {
            sum+=1.0/(n+1);
            if(sum>=x)break;
        }
        printf("%d card(s)
",n);
    }
    return 0;
}